If z = 2 + 3i,then z^{5} + (bar{z})^{5} is eq | vedclass

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(A) Given $z = 2 + 3i$,then $\bar{z} = 2 - 3i$.We need to evaluate $z^{5} + (\bar{z})^{5} = (2 + 3i)^{5} + (2 - 3i)^{5}$.Using the binomial expansion

Vedclass · Accepted Answer

$244$

Vedclass · Answer

(A) Given $z = 2 + 3i$,then $\bar{z} = 2 - 3i$.We need to evaluate $z^{5} + (\bar{z})^{5} = (2 + 3i)^{5} + (2 - 3i)^{5}$.Using the binomial expansion $(a+b)^{n} + (a-b)^{n} = 2 \sum_{k=0, 2, 4, ...} \binom{n}{k} a^{n-k} b^{k}$,we have:$z^{5} + (\bar{z})^{5} = 2 [\binom{5}{0} 2^{5} + \binom{5}{2} 2^{3} (3i)^{2} + \binom{5}{4} 2^{1} (3i)^{4}]$$= 2 [1 	imes 32 + 10 	imes 8 	imes (-9) + 5 	imes 2 	imes 81]$$= 2 [32 - 720 + 810]$$= 2 [122]$$= 244$.

If $z = 2 + 3i$,then $z^{5} + (\bar{z})^{5}$ is equal to:

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